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ECE083 - HW2 Solutions

ECE083 - HW2 Solutions - LEHIGH UNIVERSITY DEPT OF...

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Unformatted text preview: LEHIGH UNIVERSITY DEPT. OF ELECTRICAL 8: COMPUTER ENGINEERING ECE 083 INTRODUCTION TO ELECTRICAL ENGINEERING HOMEWORK # 2 @ I739. WIS- wam C’Vflwfi‘“ F340 ”If; Jam/g F5} 3% xt/éflg %Lm?65 V1) ‘12 and V3 q, @ fl: 7711.? flats/8A Mscamsm Kg W [5'41 flOSISTDI’B- 15 {503.— 171493 . MSW? Mam; Vanna-3&5 39va Fr: flxo 25v ”Ia-£345 . The circuit shown' 1n Figure P2. 6813 the dc equiv— alent of a simple residentiai power distribution "~‘ system The resistances Eabeled R1 and R2 rep ‘ I resent loads such as fights that nominally éper— @2ng £5» %W-/Mem -~-~ -~---~-~-m~--‘WWW-“M-w-"flm ate at 120 V, while R3 represents a toad such as . ' - the heating element in an even that nominally .w. ”.3965 ILi’Wwa_ 31$ "MW/wémfim ~~-~~——~-~--~-~'~~---~---~-~--.-w-—m—~w-m operates at 240 V. The resistances Eabeled Ru, ' '. represent the resistances ofwires. Rn represents ~-- ......w. ....,..,;M.. mm -~~ ~ ---«— the“neutral"wire Usemesh—curzentanalysisto determine the voltage magnitude for each load ....___n.._._.....*._,.New..." and the current in the neutral wire. ”5: WA/afica' V51» I 65" mem a 3:39 mnggflvmwm " " ‘ 1 § 5 3‘ gm)? 1:6 _, . .-. m .... ....um M. . .-.—... m M"pmw,.--_r._~.._«.._~..—pm- . -1 fihqflg-_.fiflw%mm%mfihéwxfl .-.. $1 4%: Voueflwe figura’ USM “Wm”-.. ...---WHW... ..__ -..“...me ......N _ 4?-.. «aw—www mm...;.~.._m._.1..m. 5 9 o 13 £2 .....W....-..M-..»...__'...,Wmm“..._._-,._.;...,.._~__-.W.WM.__.‘._._..mmm "WMH~-~_.5-_. . i . Mmgmgamtywwmméu 5JLl2-éSsmw— «'3 .. “g“ [6%m5W) MQWOLOL‘BKanc‘ Q____ k503i; ® (Us: 557: MM i576 TUMVLGE 4‘64 5100:; MUS: 653’ [5” 1/4 dam—cémcewmt 5M4 .3 Q mu Mag: _3_%_ z; dram/4f , WMaiLeMS [email protected] 4’54 Emmi .07.“- Was, 236 75:18 3“ m. [5- “1” >1 [Egg—”IA zumw 3 [5“;- “3 A ( fat-Kaf- L—D W 5})” Wang A} A??? MVMaJ 25- % /wfl 7/34”” 54/299726 02w; #5152946 w %%[email protected]}w5; cw . gX/kgfilfibif’fl/mflmjg‘pflflrjflh/Lflfla , . 7 r 57 7/7705” Jffl 7.7m” 7573,77 5 .7 .7 @ ,, P2.49. Soive for the node voltages shéwu in Fig» ure P249. ESQ mm v3 59 0 2.5.0 Wmfi W" ‘ N00“ mm it; ... 243 ' ‘5 ., . gfilsflzw ff: vo©+§(Q\ QQL, “(M/3 ”300‘ * P155. Soive for the values of the node voltages showr: in Figure P255. Then, find the value of ix. 20G £65}?er twp ‘. ‘ 14/4/49 é'Z’W’lfiodf , P256. Solve for the po'wer defiverad to the 169 re- sistance and for the node voitages shown in Figure P256. 7 «5Q 4W5 44,434 4674/ 74.4 m” M44" 64 4/ 6: gm: 716/!“ 6973726176147?) flag/:viofl’ea 4/@ .79 ia’axf'fif7/fi éfiém Hm 77:04.54? 6294/ Kala/c417456’ flag“ S'JWCZCK" 46F... 'Q‘ M04 ...N6u ..%.. Leia/1.4;? AAA/744.44» 44 4-; 75 . " Alhm 555;? van? .1447 Mr M12? .6 “6'41 664.65. May—wee 6/414» 666/44 __ $6“ #5612“ ix ...rfr... 'fi {+3 (/7644; éwénwnw; 5144 r “ 1,4,4 @[email protected],44w Away 7%; all @tamm/ém/é/y/ .... ,, ,. *P2. 57. Solve for $116 node voltages shown i1} Fig urE: P2. 57. P2452. Solve for the power delivered by the voltage source in Figure P262, using the mesh—current method. ' 9Q 1M ._ 2223...: ib‘of'292r'ifZé-p1‘3) :0' r7432 @ ” .. 2232222223...) 1+ . #321) 9%: c, ‘ ” " ‘ ' . - . MMQ: .%. filmy/7? .. éwmflw . (23.27 C4226? If}: QVCQ‘A1M63 . “WEN/#2 «41/: 291%: , $527. CaFv/fiichM/"CF/ag $4x9f7m% §Jufce$cflm7ry 7‘“ 2:22. W 2.477%. ../N.. )c/zm/‘VZ/LZWC’Q 7432222,”,17‘ . 227/1}: 22.213 :2: 2w . - ._ .. .. . . ,. . ' :21: —27m gmgg/éggzwé??4. _ff.*.[§..'j..jf . [.f'...'.., Lil... P168. The circuit shown in Figure P168 is the do equiv— alent of a simple residential power distribution system. The resistances labeled R; and R2 rep- resent loads such as lights that nominally Op6f- ate at 120 V, while R3 represents a load such as the heating element in an oven that nominally " ‘7 Operates at 240 V. The resistances labeled Rw represent the resistances ofwires. Rn represents the “neutral” Wire. Use mesh-current analysis to determine the voltage magnitude for each Ioa and the current in the neutral Wire. ' wam n - M931! Q). 5 fair; - L20 + 0.1 (IL—If) '1‘” 10(72;-—1"3) m.— a (fifiwfivfwce ega‘z, :27 mfl/fléfa’m’e‘o or .. i t 4, .. :Zi. 5x17"... My 7— ...! / Z72” D5, fixvfé$fl~fifiwmfmng/fls 10 Q 92 #22222222 DQDQDMAW (eigD .. - .. . . .. 0770 (23.2%; Db./.;2;Dieé2w(r;u13>D2 . c2. , . _. f; (2;nz)+ [a (2; D+5K2DJQDD=<D ...
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